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Quizzes > Science

Oxidation Half-Reaction Quiz: Which Half-Reaction Shows Oxidation?

Quick, free quiz to identify oxidized species and electron loss in redox. Instant results.

Editorial: Review CompletedCreated By: Tammy GorokhovUpdated Aug 28, 2025
Difficulty: Moderate
2-5mins
Learning OutcomesCheat Sheet
Paper art style redox reaction quiz illustration with test tubes electrons atoms and molecules on dark blue background.

This quiz helps you spot which half-reaction shows oxidation by tracking electron loss and changes in oxidation state. If you want a quick warm-up, try a chemical equation quiz, then build skills with chemical equation balancing practice, or push further with advanced equation balancing. You will get instant feedback on each item so you know what to review.

Which process describes oxidation?
Gain of electrons
Increase in mass number
Gain of protons
Loss of electrons
Oxidation is defined as the loss of electrons from an atom or ion, which increases its oxidation state. It is always paired with reduction in a redox reaction, where another species gains those electrons. Understanding this definition is fundamental to identifying redox processes.
In the reaction 2Mg + O2 ? 2MgO, which substance is the oxidizing agent?
O2
Mg
2Mg
MgO
The oxidizing agent is the species that gains electrons and is reduced. In this reaction, O2 gains electrons from Mg to form O2-, so O2 is the oxidizing agent. Mg loses electrons, becoming the reducing agent.
What is the oxidation number of chlorine in HClO3?
+1
+3
+5
+7
In HClO3, H is +1 and each O is - 2 for a total of - 6. The sum of oxidation states equals zero, so Cl must be +5. Calculating oxidation numbers is key to tracking redox changes.
Determine the oxidation state of sulfur in SO2.
-2
+6
+2
+4
Each oxygen has an oxidation state of - 2 (total - 4), so sulfur must be +4 to balance the neutral molecule. This simple calculation helps in identifying redox changes.
Which species is reduced in the reaction Zn + Cu2+ ? Zn2+ + Cu?
Zn2+
Cu2+
Zn
Cu
Reduction is the gain of electrons. Cu2+ gains two electrons to form Cu, so Cu2+ is the reduced species. Zn is oxidized, losing electrons to become Zn2+.
Which half-reaction shows the reduction of Fe3+ to Fe2+?
Fe2+ ? Fe3+ + e?
Fe3+ + e? ? Fe2+
Fe3+ + 2e? ? Fe
Fe ? Fe2+ + 2e?
Reduction means gaining electrons. The half-reaction Fe3+ + e? ? Fe2+ shows Fe3+ gaining one electron to become Fe2+. The other options either show oxidation or the wrong stoichiometry.
Balance the redox reaction in acidic solution: MnO4? + Fe2+ ? Mn2+ + Fe3+.
MnO4? + Fe2+ ? Mn2+ + Fe3+
5 Fe2+ + MnO4? + 8 H+ ? Mn2+ + 5 Fe3+ + 4 H2O
2 Fe2+ + MnO4? + 4 H+ ? Mn2+ + 2 Fe3+ + 2 H2O
MnO4? + 3 Fe2+ + 2 H+ ? Mn2+ + 3 Fe3+ + H2O
Balancing in acidic solution involves equalizing electrons and adding H+ and H2O. The correct balanced equation is 5 Fe2+ + MnO4? + 8 H+ ? Mn2+ + 5 Fe3+ + 4 H2O. This conserves mass and charge.
True or False: In all redox reactions, oxidation and reduction occur simultaneously.
True
False
Redox reactions involve the transfer of electrons, so oxidation (loss of electrons) and reduction (gain of electrons) always occur together; one cannot happen without the other. This coupling is the hallmark of redox chemistry.
Which cell notation correctly represents a galvanic cell using zinc and copper half-cells?
Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s)
Zn(s)||Zn2+(aq)|Cu2+(aq)||Cu(s)
Cu(s)|Cu2+(aq)||Zn2+(aq)|Zn(s)
Zn2+(aq)|Zn(s)||Cu(s)|Cu2+(aq)
In cell notation, the anode (oxidation) half-cell is on the left and the cathode (reduction) half-cell on the right, separated by '||'. Zn|Zn2+ is the anode, Cu2+|Cu is the cathode.
What is the standard cell potential for a cell combining Cu2+ + 2e? ? Cu (E°=+0.34 V) and Zn2+ + 2e? ? Zn (E°=?0.76 V)?
0.42 V
1.10 V
0.76 V
?0.42 V
E°cell = E°cathode ? E°anode = (+0.34 V) ? (?0.76 V) = +1.10 V. This positive value indicates a spontaneous galvanic reaction.
In a galvanic cell, electrons flow through the external circuit from the:
Cathode to anode
External circuit to cathode
Salt bridge to anode
Anode to cathode
By definition, the anode is where oxidation occurs (electrons produced) and the cathode is where reduction occurs (electrons consumed), so electrons flow from anode to cathode.
What is the proper first step when balancing redox reactions in basic solution?
Balance as if acidic, then add OH? to neutralize H+
Directly add OH? to water molecules
Use H2O and H+ only
Balance only electrons and atoms
Redox reactions in basic medium are balanced by first balancing in acid (using H+ and H2O), then adding OH? to both sides to neutralize H+. This method ensures mass and charge balance.
Which species undergoes disproportionation in the reaction 2 H2O2 ? 2 H2O + O2?
H2O2
H2O
O2
None of the above
In disproportionation, a single species is both oxidized and reduced. H2O2 is reduced to H2O and oxidized to O2, so it disproportionates.
Identify the strongest oxidizing agent from standard reduction potentials: F2 (+2.87 V), MnO4? (+1.51 V), Cl2 (+1.36 V), Fe3+ (+0.77 V).
Fe3+
F2
Cl2
MnO4?
The stronger the oxidizing agent, the more positive its reduction potential. F2 has the highest E° (+2.87 V), making it the strongest oxidizer among these.
True or False: The electrode where oxidation occurs in an electrolytic cell is called the anode.
True
False
In both galvanic and electrolytic cells, oxidation always occurs at the anode. In an electrolytic cell, the anode is connected to the positive terminal of the power source.
Calculate the reduction potential at 25°C for Ag+ + e? ? Ag when [Ag+] = 1.0×10?3 M (E° = +0.80 V).
0.88 V
0.74 V
0.62 V
0.94 V
Using the Nernst equation: E = E° ? (0.0592/n)·log(Q). Here n=1 and Q = 1/[Ag+]=10³, so E = 0.80 ? 0.0592·3 ? 0.62 V.
Determine ?G° (in kJ) for a cell with E°cell = +1.10 V involving a 2-electron transfer.
?106 kJ
?212 kJ
+316 kJ
+212 kJ
?G° = ?nFE°. With n=2 and F?96.485 kJ·V?1·mol?1, ?G° ? ?2×96.485×1.10 ? ?212 kJ. A negative ?G° indicates spontaneity.
What is the oxidation state of chromium in the dichromate ion, Cr2O7²??
+4
+2
+6
+3
Each O is ?2 (total ?14) and the overall charge is ?2, so 2Cr ? 14 = ?2 ? 2Cr = +12 ? Cr = +6. Oxidation states track electron flow.
Balance the redox reaction in basic solution: MnO4? + C2O4²? ? MnO2 + CO3²?. Which equation is correct?
3 MnO4? + C2O4²? + 4 H2O ? 3 MnO2 + 2 CO3²? + 8 OH?
MnO4? + C2O4²? ? MnO2 + CO3²?
2 MnO4? + 3 C2O4²? + 2 H2O ? 2 MnO2 + 6 CO3²? + 4 OH?
2 MnO4? + C2O4²? + 4 OH? ? 2 MnO2 + 2 CO3²? + 2 H2O
In basic medium, first balance electrons, H+ and H2O, then add OH?. The correct balanced form is 2 MnO4? + 3 C2O4²? + 2 H2O ? 2 MnO2 + 6 CO3²? + 4 OH?.
In the disproportionation of chlorine in water (Cl2 + H2O ? HCl + HOCl), what is the oxidation state of chlorine in HOCl?
?1
0
+5
+1
In HOCl, O is ?2 and H is +1, so Cl must be +1 to balance. Disproportionation involves one element in two different oxidation states.
Why is the reaction 2 NO3? ? N2O4 + O2 non-spontaneous under standard conditions?
Temperature too low
Equilibrium constant > 1
Standard cell potential is negative
High activation energy
A negative standard cell potential corresponds to a positive ?G°, indicating non-spontaneity. Kinetic factors like activation energy aren't the primary reason under standard conditions.
How many electrons are transferred in the reduction half-reaction Cr2O7²? + 14 H+ + 6 e? ? 2 Cr3+ + 7 H2O?
12
2
6
3
The given half-reaction explicitly shows 6 electrons on the reactant side. Counting electrons is essential for proper balancing of redox equations.
What term describes the extra voltage required to drive an electrolysis reaction beyond its thermodynamic potential?
Activation potential
Exchange current
Overpotential
Polarization coefficient
Overpotential is the additional voltage above the equilibrium potential needed to overcome kinetic barriers at the electrode surface. It is critical in real electrolytic systems and affects energy efficiency.
Balance the half-reaction MnO4? ? MnO2 in basic solution.
MnO4? + 2 H2O + 2 e? ? MnO2 + 4 OH?
MnO4? + 4 H+ + 3 e? ? MnO2 + 2 H2O
MnO4? + 2 H2O + 3 e? ? MnO2 + 4 OH?
MnO4? + 4 OH? + 3 e? ? MnO2 + 2 H2O
In basic solution, add H2O to balance O, H+ to balance H, then OH? to remove H+. The correct balanced form is MnO4? + 2 H2O + 3 e? ? MnO2 + 4 OH?.
Calculate the standard EMF for the cell combining Ag+ + e? ? Ag (E°=+0.80 V) and Sn2+ + 2e? ? Sn (E°=?0.14 V).
1.14 V
0.66 V
0.14 V
0.94 V
E°cell = E°cathode ? E°anode. Ag+/Ag has the higher reduction potential (+0.80 V) and acts as cathode; Sn2+/Sn is anode (?0.14 V). Thus E°cell = 0.80 ? (?0.14) = +0.94 V.
0
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Study Outcomes

  1. Understand Electron Transfer Mechanisms -

    Grasp how electrons move between species in redox reactions, distinguishing oxidation from reduction events.

  2. Identify Oxidation States and Redox Pairs -

    Determine oxidation numbers for elements in inorganic compounds and pinpoint which species act as oxidizing or reducing agents.

  3. Construct Half-Reactions -

    Write balanced half-reactions for both oxidation and reduction processes to clearly represent electron flow.

  4. Balance Full Redox Equations -

    Apply the half-reaction method in acidic or basic conditions to achieve mass and charge balance in overall redox equations.

  5. Analyze Reaction Scenarios -

    Assess various inorganic chemistry examples to predict products and verify electron balance in redox processes.

  6. Evaluate Quiz Performance with Feedback -

    Use instant feedback to identify areas of strength and target common pitfalls in redox reaction questions.

Cheat Sheet

  1. Assign Oxidation Numbers -

    Master the IUPAC rules for assigning oxidation states: free elements are zero, monatomic ions equal their charge, and common patterns (O is - 2, H is +1) guide you. Practice on molecules like H₂O (H = +1, O = - 2) and MnO₄❻ (Mn = +7) to build confidence.

  2. Identify Oxidation and Reduction -

    Use the mnemonic "OIL RIG" (Oxidation Is Loss, Reduction Is Gain) to track electron flow; oxidation increases oxidation number, reduction decreases it. For example, in Zn + Cu²❺ → Zn²❺ + Cu, Zn loses electrons (oxidized) and Cu²❺ gains them (reduced).

  3. Balance Half-Reactions -

    Split redox equations into oxidation and reduction half-reactions and balance atoms (add H₂O, H❺ or OH❻) before electrons. Finally, equalize electron transfer between halves - this systematic approach from university inorganic texts ensures precise balancing even in acidic or basic media.

  4. Calculate Cell Potentials -

    Use standard reduction potentials (E°) from trusted tables to find E°cell = E°cathode - E°anode; positive E°cell indicates spontaneity. Relate ΔG° = - nFE°cell to predict reaction feasibility in galvanic and electrolytic cells.

  5. Apply to Titrations and Industry -

    Practice redox titrations (e.g., KMnO₄ vs Fe²❺) to sharpen endpoint recognition and concentration calculations. Explore real-world processes like electroplating and corrosion prevention to see redox principles in action and deepen conceptual understanding.

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